MARKING SCHEME- CHEMISTRY (043)

SAMPLE PAPER (CLASS - XII)

Q.No.

Value Points Marks

1. T1 < T2 < T3 1 2. XeF2 1 3.

It strengthens the bond between CO and the metal.

1

4. The oxidation state of P in PCl5 is +5 it cannot increase its oxidation state beyond +5 but it can decrease from +5 to +3.

1

5. Iodobenzene 1 6. Schottky defect

It is shown by ionic substances in which the cation and anion are of almost similar sizes. / ionic substances having high coordination number.

1 1

7. The plot is nearly a straight line and can be extrapolated to zero concentration(i.e. from the intercept) to find the value of 0

m = 150.0 S cm2 mol-1

[Λm

c = Λm0 - A c ]

A = - slope = xy

= 034.0

0.1470.150

= 88.23 S cm2 mol-1

1 ½ ½

8. r = 125 pm, a = ?

for fcc structure 22

ar

414.12125 xxa = 353.5 pm

½ ½ ½ +

½

9. (i) (ii)

1 1

(i) (ii)

OR 3 Cl2 + 6 NaOH → 5 NaCl + NaClO3 + 3 H2O 2 Fe3+ + SO2 + 2 H2O → 2 Fe2+ + SO4

2- + 4 H+

( ½ mark to be deducted for an unbalanced chemical equation)

1 1

10.

KCl is an electrolyte, it undergoes dissociation

mKT

i

mKiT

f

f

ff

fT = 0 –(-0.24) = 0.24 0

Molar mass of KCl = 39 + 35.5 = 74.5 u substituting the values

10005.086.11005.7424.0

xxxxi

i = 1.92 α is the degree of dissociation i = 1 + α α =1.92 – 1 = 0.92 Percentage dissociation = 92 % (or any other suitable method)

½ ½ ½ ½

11. (i) (ii)

Hydraulic washing: Principle involved: differences in gravities of the ore and the gangue particles e.g.oxide ores ( haematite), native ores Au, Ag (any one example) Zone refining: Principle involved: the impurities are more soluble in the melt than in the solid state of the metal. e.g. germanium, silicon, boron, gallium and indium (any one example)

1 ½ 1 ½

12.

(i) (ii)

A warm solution is obtained on mixing the two liquids A and B indicate that the process of mixing is exothermic. ( ∆H mixing = - ve). So the solution shows a negative deviation from the Raoult’s law.

The forces of interaction between A and B molecules are more than in the A-A and B-B molecules. So the partial vapour pressure of each component will be less and the partial vapour pressure of the solution will also be less than that from the Raoult’s law. For CH3OH ( non electrolyte) i = 0 KCl → K+ + Cl- i = 2 Na3PO4 → 3 Na+ + PO4

3- i = 4 CH3OH < KCl < Na3PO4

½ 1 ½ 1

13.

Nernst equation:

][][log

20591.0

2

20

CuMgEE cellcell

]10[]10[log

20591.071.2 4

3

cellE

Ecell = 2.68 V Ecell increases with the increase in the concentration of Cu2+ ions and decrease in the concentration of Mg2+ ions.

½ ½ ½ + ½ 1

14. (i) (ii) (iii)

Decomposition of ozone into oxygen is an exothermic process (∆H = -ve) and results in an increase in entropy(∆S = +ve), resulting in large negative Gibbs energy change(∆G = -ve). Decomposition of ozone to oxygen is a spontaneous process. As we move down the group, the size of the central atom increases and the electronegativity decreases. therefore the bond pair of electrons lie away from the central atom, the force of repulsion between the adjacent bond pairs decreases. So the bond angle decreases. The bleaching action of Cl2 is due to the oxidation of coloured substance to colourless by nascent oxygen. Cl2 + H2O → 2 HCl + [O] Nascent oxygen

1 1 ½ ½

15. (i) (ii) (iii)

Mn has the configuration 3d54s2. Hence the configuration of Mn2+ in [MnBr4]2- is 3d5. Mn2+

↑ ↑ ↑ ↑ ↑ 3d 4s 4p Since it is tetrahedral in shape, the hybridization is sp3. There are five unpaired electrons.

[Pt(NH3)BrCl(NO2)]-

½ ½ (½ +½) 1

16. (i) (ii) (iii)

When silver nitrate solution is added to potassium iodide solution, a precipitate of silver iodide is formed which adsorbs iodide ions from the dispersion medium and a negatively charged colloidal solution is formed. As the size of the gold sol particles increases, the colour of the solution changes from red to purple, then blue and finally golden because the colour of colloidal solution depends on the wavelength of the light scattered by the dispersed particles and wavelength further depends on the size of the particles. When oppositely charged sols are mixed in almost equal proportions, neutralization of their charges occur and precipitation occurs.

1 1 1

17. (i) (ii)

Cl

Undergoes SN1 faster. It is a secondary halide, which forms a secondary carbocation which is more stable than the primary carbocation so greater will be its ease of formation from the corresponding alkyl halide and faster will be the rate of reaction. The para-isomer

½ ½

(iii)

ClCl

has the highest melting point as compared to their ortho- and meta- isomers. The para-isomer is more symmetrical and fits into the crystal lattice better, as a result intermolecular forces are stronger, higher temperature required to melt the para-isomer. CH2Cl2 < CHCl3 < CCl4 ( increasing order of density) The density increases with the increase in the number of the halogen atoms.

½ ½ ½ ½

18. A (C7H6O) NaOH B (C7H8O) + C (sodium salt of an acid)

[O]

NaOH + CaO

A D (A) undergoes disproportionation in presence of an alkali (Cannizzaro reaction) so there is no α hydrogen. (C) undergoes decarboxylation. A

CHO

B

CH2OH

C

COONa

D

CHO CH2OH

NaOH+

COONa

½ ½ ½ ½ ½

A B C

COONa

NaOH + CaO

C D

½

19. (a) (i) (ii) (b)

Chemical test Methylamine

(10 aliphatic amine) Dimethylamine (20 aliphatic amine)

Carbylamine test: To 1 ml of the organic compound add an alcoholic solution of KOH and CHCl3

Gives offensive smell.

No odour obtained (no reaction)

Chemical test Aniline

(10 aromatic amine) Benzylamine (10 aliphatic amine)

Azo dye test: To 1mL of the organic compound add HNO2 (NaNO2 + dil. HCl) at 273-278 K. then add an alkaline solution of β naphathol to the solution.

A brilliant red dye is obtained

No dye obtained

( any other suitable test) Four structural isomers are possible for C3H9N Only primary amines react with HNO2 to liberate nitrogen gas. 10 amines: CH3CH2CH2NH2

CH3-CH-CH3

NH2

No reaction for 20 and 30 amines 20 amine: CH3-NH-C2H5

30 amine:

CH3-N-CH3

CH3

1 1 ½ ½

20. (a) (i)

When acid chloride is hydrogenated over catalyst, palladium on barium sulphate they are reduced to the corresponding aldehydes. This reaction is called Rosenmund reduction.

½

(ii) (b) (a) (i) (ii) (b)

COCl

H2

Pd - BaSO4

CHO

Benzoyl chloride Benzaldehyde Alkali metal salts of carboxylic acids undergo decarboxylation on electrolysis of their aqueous solutions and form hydrocarbons having twice the number of carbon atoms present in the alkyl group of the acid. This reaction is called Kolbe electrolysis reaction. 2CH3COOK + 2H2O → CH3-CH3 + 2CO2 + 2KOH + H2 Acetaldehyde : CH3CHO Acetone : CH3COCH3 Di-tert-butyl ketone : (CH3)3C-CO-C(CH3)3 Di-tert-butyl ketone < Acetone < Acetaldehyde OR

(i) Cl2 / Red phosphorous

(ii) H2OCH3-CH2-COOH CH3-CH-COOH

Cl

CH3

+ CrO2Cl2(i) CS2

(ii) H3O+

CHO

Benzoic acid

COOH

4-Nitrobenzoic acid

COOHO2N

4-Methoxybenzoic acid

½ ½ ½ 1 1 1

COOHCH3O

4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid

1

21. (i) (ii) (iii)

Buta-1,3diene : CH2=CH-CH=CH2 and Acrylonitrile : CH2=CH(CN) (either name or structure) Neoprene is classified as an Elastomer. (the polymeric chain are held together by weakest intermolecular forces) Yes a co-polymer can be formed in addition and condensation polymerization.

½ ½ 1 1

22. (i) (ii) (iii)

Keratin is insoluble in water. It is a fibrous protein in which the polypeptide chains are held together by strong intermolecular forces, hence insoluble in water. α-D-Glucopyranose

4

5O

1

23

H

HOH

OHOH

HH

OH H

CH2OH6

The sequence in the complimentary strand is ATGCTTGA

½ ½ 1 1

23. (a) (i) (ii)

No the chemist did not give the appropriate medicine. Cimetidine is an antihistamine, but it is an antacid and not an antiallergic drug.Antacid and antiallergic drugs work on different receptors. Therefore cimetidine cannot be used to treat nasal congestion. Critical thinking Social responsibility (or any other two reasons)

½ 1 ½ 1 1

24. (a)

Mechanism of hydration of ethene to ethanol by acid catalysed hydration:

(b) (i) (ii) (c) (a)

CH3CH2CH2-OH

Heat

SOH 42 CH3CH=CH2 HBr CH3-CH(Br)-CH3

Propanol aq. KOH CH3-CH(OH)-CH3 Propan-2-ol CH3CH2CH2-OH 2SOCl CH3CH2CH2-Cl ↓NaOH CH3CH2CH2-ONa CH3CH2CH2-Cl + CH3CH2CH2-ONa → CH3CH2CH2-O-CH2CH2CH3

OH

conc. HNO3

OH

NO2

NO2

O2N

2,4,6-Trinitrophenol OR The mechanism of the reaction of HI with methoxymethane involves the following steps: Step I : protonation of ether molecule

1 ½ ½ 1 ½ ½ ½ ½

(b) (i) (ii) (c)

Step II : nucleophilic attack by I- by SN2 mechanism

Step III : when HI is in excess and the reaction is carried out at high temperature, methanol formed in the second step reacts with another molecule of HI and is converted to methyl iodide.

Methyl iodide A B

ONa

OH

COOH

A : CH3CHO B : CH3-CH(OH)-CH3

O2N

OC2H5OC2H5 conc. HNO3

conc. H2SO4

1- Ethoxy-4-nitrobenzene

½ ½ 1 ½ + ½ ½ ½ ½ ½

25. (a)

A : MnO2 B : K2MnO4 C : KMnO4

½ ½ ½

(b) (i) (ii) (iii) (a) (i) (ii) (b) (i) (ii) (iii)

2 KMnO4 K2MnO4 + MnO2 +O2

Mn3+ and Co3+ are the strongest oxidizing agents from the data given. Copper (I) compounds are unstable in aqueous solution and undergo disproportionation 2 Cu+ → Cu2+ + Cu Cu2+ (aq) is more stable than Cu+ because it has high negative ∆hydrH0as compared to Cu+.

The highest oxidation state of a metal is exhibited in its oxide as oxygen has the ability to form multiple bonds to metal atoms. OR 2 Cu2+ + 4I- → Cu2I2 (s) + I2

Cr2O7

2- + 14 H+ + 6 Fe2+ → 2 Cr3+ + 6 Fe3+ 7 H2O The third series (5d elements) of the transition elements have the highest first ionization enthalpy ( Hf to Au). This is because of the poor shielding of the nucleus by the 4f electrons in the 5d elements which results in greater effective nuclear charge on the valence electrons. Due to the lanthanoid contraction the change in the ionic radii in the lanthanoids is very small, their chemical properties are similar, so the separation is difficult. +3 is the most stable oxidation state of lanthanides. Ions in +2 tend to change to +3 by losing electrons so act as reducing agents, whereas ios in +4 tend to change to +3 by gaining electrons so act as oxidizing agents.

½ ½ +½ ½ ½ 1 1 1 ½ ½ 1 1

26. (i)

For a chemical reaction with rise in temperature by 100, the rate constant is nearly doubled.

(T2 = T1 + 100) Increasing the temperature of the substance increases the fraction of molecules which collide with energies greater than Ea (activation energy) As the temperature increases (T2) , the fraction of molecules having energy equal to or greater than activation energy gets doubled leading to doubling the rate of reaction.

½ 1 1

(ii) (i) (ii)

k = A e-Ea/ RT (Arrhenius equation) k = A e-Ea/ RT

TK

RTEa 28000

Ea= 28000 K x 8.314 J / K / mol Ea = 232.79 kJ / mol OR t 99 % = 2 t 90 % Given for reactant Q. for first order reaction

kx

kaaa

kt 303.2210log303.2

99.0log303.2 2

%99

kkaaa

kt 303.210log303.2

90.0log303.2

%90

Therefore t 99 % = 2 t 90 % , therefore it is a first order reaction with respect to Q From the graph it is evident that the concentration of R decreases linearly with time, therefore the order with respect to R is zero. The overall order of the reaction is 1. Units for rate constant = s-1 Rate = k [Q]1 [R]0 For a first order reaction

][][log303.2 0

RR

kt

Time required for the 3 /4 th of the reaction

[R]0 = a, [R] = aaa41

43

a

ak

t

41log303.2

43

4log1054.2

303.23

43

xt

= 545 s

½ 1 ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ ½